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CHARACTERIZING THE ORBITS OF THE ROTATION MAP
MEGAN ADAMO
Abstract. We begin by developing the necessary tools to analyze subsets of
topological spaces. We then define a topological space and compare continuity
in the topological sense to the traditional − δ method. Lastly, we use the
tools we have developed to characterize the orbits of the rotation map on the
circle.
Contents
1. Introduction
2. Bounds, Limit Points, and Openness
3. Topological Spaces
4. Continuity
5. Orbits of the Rotation Map
Acknowledgments
References
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1. Introduction
In this paper, we study the rotation map on the circle, an important example in
the theory of dynamical systems. In order to study the rotation map we must first
develop basic topological concepts, including bounds, limit points, openness, and
continuity. We then apply these concepts to study the orbits of the rotation map,
which, for a given angle θ, rotates all the points in the circle by θ. We show that
when the angle is an irrational multiple of 2π, the orbits are dense, and when the
angle is a rational multiple of 2π, they are finite.
The organization of the paper is as follows: in Section 2, we discuss metric spaces,
upper and lower bounds, limit points and some applications, and open sets and their
properties. In Section 3, we define a topological space and give examples of different
topological spaces. In Section 4, we examine the − δ definition of continuity of a
real-valued function on R and explore how it relates to the topological definition of
continuity. We conclude in Section 5 by studying the orbits of the rotation map.
2. Bounds, Limit Points, and Openness
We begin by examining bounds on subsets of ordered sets. Next, we define a
metric space and examine limit points, which we then use to examine openness in
sets.
Date: November 17, 2015.
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MEGAN ADAMO
Definition 1. An ordered set S is a set which has an order relation, <, satisfying
the following two properties:
1. If x ∈ S and y ∈ S, then exactly one of the following is true:
x < y, y < x, or y = x.
2. (Transitivity) If x, y, z ∈ S, then x < y and y < z implies x < z.
Definition 2. Let M be an ordered set, and A ⊂ M. An element β ∈ M is an
upper bound for A if for all x ∈ A, x ≤ β. A is called bounded above if it has an
upper bound. An upper bound β is called the least upper bound or supremum if
for any α which is an upper bound, β ≤ α.
Definition 3. Let N be an ordered set, and B ⊂ N. An element α ∈ N is a
lower bound for B if for all x ∈ B, x ≥ α. B is called bounded below if it has a lower
bound. A lower bound α is called the greatest lower bound or infimum if for any γ
which is a lower bound, α ≥ γ.
Claim 1. If E is a nonempty subset of an ordered set, X, α ∈ X is a lower bound
of E and β ∈ X is an upper bound of E, then α ≤ β.
Proof. Let α ∈ X be a lower bound of E. E is nonempty, so let x ∈ E. By definition
of a lower bound, α ≤ x. Let β be an upper bound of E. By definition of an upper
bound, x ≤ β. By transitivity, α ≤ β.
Definition 4. Take a set S with its elements called points. S is a metric space
if any two points x and y in S have a real number distance associated with them.
The real number distance between x and y is written as d(x, y) and satisfies the
following three properties:
1. d(x, y) > 0 if x 6= y, and d(x, x) = 0.
2. d(x, y) = d(y, x).
3. d(x, y) ≤ d(x, z) + d(y, z) for any z ∈ S.
Example 1. An example of a metric space is the real line with standard distance.
Take x, y ∈ R1 . The distance is defined by d(x, y) = |x − y| where | · | is the absolute
value on R.
Example 2. Another example of a metric space is arclength on S 1 . Take x, y ∈ S 1 ,
with x = eiθ and y = eiφ . The arclength is defined by d(x, y) = |θ − φ|, where θ
and φ are in radians.
Definition 5. Let B (p) denote the points x around a point p in a metric space S
such that d(x, p) < . A point p ∈ S is a limit point of a subset A of S if for any
> 0, there exists some q ∈ A such that q ∈ B (p) and q 6= p.
Definition 6. A subset X of a metric space Y is dense in Y if for all x ∈ X, x ∈ Y
or x is a limit point of Y.
Example 3. We can construct a bounded set of real numbers with one limit point.
Proof. Let
1
S = { |n ∈ N − {0}}.
n
CHARACTERIZING THE ORBITS OF THE ROTATION MAP
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Let > 0, ∈ R. By the Archimedean Property, we know that given > 0, there
exists an n such that n1 < . Because n1 ∈ S and n1 6= 0, 0 is a limit point of S.
We can show that 0 is also the only limit point of S. Let x ∈ R, x 6= 0.
Case 1. x < 0. Set = −x
2 . Then B (x) ∩ S = ∅, so x is not a limit point.
Case 2. x > 1. Set = x2 . Then B (x) ∩ S = ∅, so x is not a limit point.
Case 3. x = 1. Set = 31 . Then B (x) ∩ S = ∅, so x is not a limit point.
Case 4. 0 < x < 1. Set = 21 (x − p1 ), where p1 < x. Then B (x) ∩ S = ∅ or
B (x) ∩ S = n so x is not a limit point.
Theorem 2.1. Let Si , i = 1, ...n be subsets of a metric space
Sn X. If Xi isSthe set of
limit points of Si , and X is the set of limit points of S = i=1 Si , then Xi = X.
S
Proof. We first show that Xi ⊂ X. Let x ∈ Xi , so x is a limit point of Si . Take
an neighborhood B (x) of x. Since x is S
a limit point of Si , there exists si ∈ Si
n
such that si 6= x and si ∈ S
B (x). So si ∈ i=1 Si implies si ∈ S. Thus x is a limit
point of S, so x ∈ X, and X
Si ⊂ X.
S
We now show that X ⊂ Xi . Take x such that x ∈
/
Xi . Since x is not a
limit point of Si , for each i there exists i such that Bi (x) ∩ Si = ∅ or {x}. Let
= min(1 , ..., n ). So B (x) ⊂ B1 (x) for all i. Assume
S by contradiction that there
exists s ∈ S, s 6= x such that s ∈ B (x). Since s ∈ Si , there exists some i such
that there exists s ∈ Si . But we know that B (x) ∩ Si = ∅ or {x} because B (x) ∩ Si
is contained in Bi (X) ∩ Si . Because s 6= x, such an s does not exist.
Example 4. Theorem 2.1 can fail for an infinite union of sets: Let LP
S (X) denote
the
set
of
limit
points
of
the
set
X.
Let
{U
}
be
such
that
LP
(
α
α Uα ) 6=
S
α LP (Uα ).
Proof. Let Uα = {α}, α ∈ R. Then Uα is a finite set and thus has no limit points.
So
∪α∈R LP ({α}) = ∪α∈R ∅ = ∅.
However, ∪α∈R {α} = R, so LP (∪α∈R {α}) = R. Thus LP (∪α∈R {α}) 6= ∪α LP ({α}).
Definition 7. U ⊆ M, where M is a metric space, is open if for all x ∈ U, there
exists > 0 such that B (x) ⊂ U.
Theorem 2.2. If U and V are open subsets of a metric space, then U ∩ V is open.
Proof. Case 1. If U ∩ V = ∅, then U ∩ V is open because the empty set is an open
set. (The empty set is an open set because there are no x in the empty set, thus
the condition for all x ∈ U, there exists > 0 such that B (x) ⊂ U is vacuously
satisfied.)
Case 2. U ∩ V 6= ∅. Take x ∈ U ∩ V. Because U and V are open sets, there exists
U > 0 and V > 0 such that BU (x) ⊂ U and BV (x) ⊂ V. Thus we can choose
= min(U , V ) > 0 such that B (x) ⊂ U ∩ V.
Tn
Corollary 2.3. If U1 , ..., Un are open sets in a metric space, then i=1 U is open.
Example
5. However, the infinite intersection of open sets is not necessarily open:
T∞
1
The set n=1 ( −1
n , n ) is not open.
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MEGAN ADAMO
T∞
1
Proof. n=1 ( −1
n , n ) = {0} and a set consisting of a single point is not open. We
know that 0 is the result of the intersection, because given any other point r, we
can choose an n such that n1 < |r|.
S
Theorem 2.4. If {Ui }i∈I are open, then Ui is open.
S
Proof. Take x ∈ i Ui . For some i, x ∈ Ui . Because S
Ui is open, there exists > 0
such that B (x) ⊂ Ui . Because B (x) ⊂ Ui , B (x) ⊂ i Ui .
3. Topological Spaces
We begin by defining a topological space. We then discuss examples of different
topological spaces and show how they satisfy the basic axioms.
Definition 8. A topological space X is a set X with a collection T of subsets of
X called open sets satisfying the following:
1. ∅, X ∈ T. (Thus ∅ and X are both open.)
2. If {Uα } is a collection of open sets (Uα ∈ T ), then ∪α Uα is open.
3. If U1 , ..., Un are open, so is U1 ∩ U2 ... ∩ Un .
Example 6. If X is any set, there are two easily defined topologies on X :
1. The indiscrete topology, where the only open sets are ∅ and X.
2. The discrete topology, where everything is open: T = PX.
Example 7. A metric space M is a topological space with open sets defined as in
the previous section.
We must show that the three axioms are satisfied.
1. Noted in the proof of Theorem 2.2.
2. Proved in Theorem 2.4.
3. Proved in Corollary 2.3.
Example 8. Let X be a set. The open sets of the cofinite topology on U are
T = {∅} ∪ {U | X − U is finite}. This is a topology.
Proof. We must show the three axioms are satisfied.
1. ∅ is open in X by definition of the cofinite topology. X is open in X because
X − X = ∅ and ∅ is finite.
2. Arbitrary U nion.
Case 1 :SUβ 6= ∅ for some β. Because Uβ is open,
S we know that X − Uβ is finite.
Since X − α Uα ⊂ X − Uβ , we know that X − α Uα is finite because a subset of
a finite set is necessarily finite.
S
Case 2 : Uα = ∅ for all α. Then Uα = ∅. By definition, ∅ is open, so this case
of the infinite union is also open.
3. F inite Intersection. Case 1 : Ui 6= ∅ for all i ≤ n.
n
n
[
[
c
X − ∩U1 ∩ · · · ∩ Un = (∩U1 ∩ · · · ∩ Un ) =
(X − Ui ) =
Ui c .
i=1
c
i=1
Tn
Because Ui is open, Ui is finite by definition, so X − i=1 Ui is open. Thus the
finite intersection of open sets is open.
T
Case 2 : Un = ∅ for all n. Then n Un = ∅. By definition ∅ is open, thus this
case of the finite intersection is also open.
Remark 1. If X is finite, the cofinite topology is the discrete topology.
CHARACTERIZING THE ORBITS OF THE ROTATION MAP
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4. Continuity
We examine two definitions of continuity and prove that they are equivalent.
Definition 9 (The − δ definition of continuity). If f is an arbitrary metric space
we say f is continuous if for all x ∈ R and for all > 0 there exists some δ = δ(x, )
so that
d(x, y) < δ =⇒ d(f (x), f (y)) < .
Definition 10 (topological definition of continuity). Let X and Y be topological
spaces and let f : X → Y. Then f is continuous if for all open U ⊂ Y, f −1 (U ) is
open in X.
Theorem 4.1. The topological definition of continuity is equivalent to the − δ
definition of continuity if f : A → R is a function on an arbitrary metric space.
Proof. Topological definition =⇒ − δ definition. Suppose f : A → R is a
function on an arbitrary metric space and is continuous in the topological sense.
Let x ∈ R, > 0. Then f −1 (B (f (x))) is open. Because f −1 (B (f (x))) is open and
x ∈ f −1 (B (f (x))), there exists δ > 0 so that Bδ (x) ⊂ f −1 (B (f (x))). Therefore
f (Bδ (x)) ⊂ B (f (x)). This shows that if
d(x, y) < δ, then d(f (x), f (y)) < .
− δ definition =⇒ topological definition. We want to show that if U is
open, then f −1 (U ) is open. Let x be any element of f −1 (U ). Because U is open,
there exists > 0 so that B (f (x)) ⊂ U. We know there exists δ > 0 such that if
d(x, y) < δ, then d(f (x), f (y)) < . In particular, this says that Bδ (x) ⊂ f −1 (U ).
5. Orbits of the Rotation Map
Definition 11. A dynamical system is a set S and a map f : S → S.
Definition 12. Take a point p ∈ S. Then {f (p), f (f (p)), ..., f n (p)...} is the orbit
of p under f.
Let the notation {α} denote the fractional portion of α. Let the notation bαc
denote the largest integer less than or equal to α. So {α} = α−bαc, and {α} ∈ [0, 1).
Theorem 5.1 (Dirichlet approximation theorem). If α ∈ R, n ∈ N then there exist
p, q ∈ Z such that |pα − q| < n1 .
Proof. Split up [0, 1] into n pieces, so each piece is size n1 .
Now consider the real numbers {α}, {2α}, ..., {nα}, {(n+1)α}. By the Pigeonhole
Principle,
r r+1 there exist some i, j, r such that {iα} and {jα} lie in the same interval
. Thus
n, n
1
|{iα} − {jα}| < .
n
1
|iα − biαc − (jα − bjαc)| < .
n
1
|(i − j)α − biαc + bjαc| < .
n
Let q = bjαc − biαc and p = i − j. Thus we have |pα − q| < n1 and we are done.
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MEGAN ADAMO
An important example of a dynamical system is the following problem regarding
the orbits of the rotation map.
Theorem 5.2. For θ ∈ [0, 2π] , let Rθ be the map from the circle to itself given by
counterclockwise rotation by angle θ. Consider the orbit of 1 under Rθ . The orbit
of 1 is finite if θ ∈ 2π · Q and dense in S 1 otherwise.
Proof. Suppose the orbit, Iθ = {1, Rθ (1), ..., Rθ n (1)}, is finite. We know that
Rθ (1) = eiθ . By the definition of orbit, Rθn+1 (1) ∈ Iθ . Consider Rθn+1 (1) = Rθm (1)
for some 1 ≤ m ≤ n. Then we have that
e(n+1)iθ = eimθ
e(n+1)iθ
= ei(n+1−m)θ
1=
eimθ
2kπ = (n + 1 − m)θ
2kπ
θ=
n+1−m
Therefore if Iθ is finite, θ must be a rational multiple of 2π.
Conversely, if θ is a rational multiple of 2π then the orbit is finite. Given that
θ = 2πp
q where p, q are relatively prime, we can define
p
Iθ = {Rθ (1) = ei2π q , Rθ2 (1) = ei2π
2p
q
...Rθq (1) = ei2π
qp
q
= e2πip = 1}.
Thus the orbit is finite and contains q elements.
If θ is not a rational multiple of π, then we want to prove that Iθ = {e2πinθ | n ∈
N} is dense in the unit circle.
θ
∈ [0, 1). Let p ∈ S 1 , p = eiφ with
Let θ ∈ [0, 2π) be irrational. Define r = 2π
φ
φ ∈ [0, 2π]. Define β = 2π so β ∈ [0, 1]. Pick M such that 2π
M < . Dirichlet’s
θ
1
Approximation Theorem implies that there exist p, q such that |q 2π
− p| < M
. This
implies that
2π
|qθ − 2πp| <
< .
M
So we know that Rθq (1) is within of 1. Thus the points Rθq (1), Rθ2q (1), ..., RθM (1)
break S 1 into M pieces of length less than . Then the distance from β to any point
in Iθ is less than , so every point in the unit circle is a limit point of the irrational
orbit, so Iθ is dense in the unit circle.
Remark 2. If θ = pπ
q , where p, q ∈ Q, then the set I has exactly q elements if
q
gcd(p, q) = 1 and gcd(p,q)
if gcd(p, q) 6= 1.
Remark 3. If e2πiα is another point on S 1 , then the orbit of e2πiα under Rθ is
just the orbit of 1 rotated by 2πα.
Acknowledgments. It is a pleasure to thank my mentors, Sean Howe and Abigail
Ward, for their thoughtful guidance and immense patience throughout the entire
process. I would also like to thank Peter May and Làszlò Babai for making the
entire program a great experience. Additional gratitude is extended to Peter Haine
and Harris Daniels for their help in the editing process.
CHARACTERIZING THE ORBITS OF THE ROTATION MAP
References
[1] James R. Munkres. Topology, Second Edition. Prentice Hall. 2000.
[2] Stephen Abbott. Understanding Analysis, Second Edition. Springer Science + Media. 2001.
[3] Walter Rudin. Principles of Mathematical Analysis. McGraw-Hill, Inc. 1976
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